If $\lim _ { x \rightarrow 0 } \frac { a x - \left( e ^ { 4 x } - 1 \right) } { a x \left( e ^ { 4 x } - 1 \right) }$ exists and is equal to $b$, then the value of $a - 2 b$ is $\underline{\hspace{1cm}}$.

If $\lim _ { x \rightarrow 0 } \frac { a e ^ { x } - b \cos x + c e ^ { - x } } { x \sin x } = 2$, then $a + b + c$ is equal to $\_\_\_\_$.

If $\lim _ { x \rightarrow 0 } \left[ \frac { \alpha x e ^ { x } - \beta \log _ { e } ( 1 + x ) + \gamma x ^ { 2 } e ^ { - x } } { x \sin ^ { 2 } x } \right] = 10 , \alpha , \beta , \gamma \in R$, then the value of $\alpha + \beta + \gamma$ is $\underline{\hspace{1cm}}$.

If $t = \sqrt { x } + 4$, then $\left( \frac { \mathrm { d } x } { \mathrm {~d} t } \right) _ { t = 4 }$ is:
(1) 4
(2) Zero
(3) 8
(4) 16

Let $f(x)$ be a polynomial function such that $f(x) + f ^ { \prime } (x) + f ^ { \prime \prime } (x) = x ^ { 5 } + 64$. Then, the value of $\lim _ { x \rightarrow 1 } \frac { f(x) } { x - 1 }$ is equal to
(1) $- 15$
(2) $15$
(3) $- 60$
(4) $60$

$\lim _ { x \rightarrow 0 } \frac { \cos ( \sin x ) - \cos x } { x ^ { 4 } }$ is equal to
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 12 }$

$\lim_{x \to \frac{\pi}{2}} \tan^2 x \left[(2\sin^2 x + 3\sin x + 4)^{\frac{1}{2}} - (\sin^2 x + 6\sin x + 2)^{\frac{1}{2}}\right]$ is equal to
(1) $\frac{1}{12}$
(2) $-\frac{1}{18}$
(3) $-\frac{1}{12}$
(4) $\frac{1}{6}$

$\lim_{x \rightarrow \frac{\pi}{4}} \frac{8\sqrt{2} - (\cos x + \sin x)^7}{\sqrt{2} - \sqrt{2}\sin 2x}$ is equal to
(1) 14
(2) 7
(3) $14\sqrt{2}$
(4) $7\sqrt{2}$

$f , g : R \rightarrow R$ be two real valued function defined as $f ( x ) = \left\{ \begin{array} { c l } - | x + 3 | & , x < 0 \\ e ^ { x } & , x \geq 0 \end{array} \right.$ and $g ( x ) = \left\{ \begin{array} { l l } x ^ { 2 } + k _ { 1 } x , & x < 0 \\ 4 x + k _ { 2 } , & x \geq 0 \end{array} \right.$, where $k _ { 1 }$ and $k _ { 2 }$ are real constants. If $gof$ is differentiable at $x = 0$, then $gof ( - 4 ) + gof ( 4 )$ is equal to
(1) $4 \left( e ^ { 4 } + 1 \right)$
(2) $2 \left( 2 e ^ { 4 } + 1 \right)$
(3) $4 e ^ { 4 }$
(4) $2 \left( 2 e ^ { 4 } - 1 \right)$

Let $x = 2$ be a root of the equation $x ^ { 2 } + p x + q = 0$ and $f ( x ) = \left\{ \begin{array} { c l } \frac { 1 - \cos \left( x ^ { 2 } - 4 p x + q ^ { 2 } + 8 q + 16 \right) } { ( x - 2 p ) ^ { 4 } } , & x \neq 2 p \\ 0 , & x = 2 p \end{array} \right.$. Then $\lim _ { x \rightarrow 2 p ^ { + } } [ f ( x ) ]$ where $[ \cdot ]$ denotes greatest integer function, is
(1) 2
(2) 1
(3) 0
(4) $- 1$

$\lim _ { x \rightarrow 0 } \left( \left( \frac { 1 - \cos ^ { 2 } ( 3 x ) } { \cos ^ { 3 } ( 4 x ) } \right) \left( \frac { \sin ^ { 3 } ( 4 x ) } { \left( \log _ { e } ( 2 x + 1 ) \right) ^ { 5 } } \right) \right)$ is equal to
(1) 15
(2) 9
(3) 18
(4) 24

Let $y = f(x) = \sin^3\left(\frac{\pi}{3}\cos\left(\frac{\pi}{3\sqrt{2}}\left(-4x^3 + 5x^2 + 1\right)^{3/2}\right)\right)$. Then, at $x = 1$,
(1) $2y' + \sqrt{3}\pi^2 y = 0$
(2) $2y' + 3\pi^2 y = 0$
(3) $\sqrt{2}y' - 3\pi^2 y = 0$
(4) $y' + 3\pi^2 y = 0$

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f\left(\frac{\pi}{2}\right) = 0$ and $f'\left(\frac{\pi}{2}\right) = 1$ and let $g(x) = \int_x^{\pi/4} (f'(t)\sec t + \tan t \cdot f(t)\sec t)\,dt$. Then $\lim_{x \to \pi/2} \frac{g(x)}{(x - \pi/2)^2}$ is equal to $\_\_\_\_$.

Let $y ( x ) = ( 1 + x ) \left( 1 + x ^ { 2 } \right) \left( 1 + x ^ { 4 } \right) \left( 1 + x ^ { 8 } \right) \left( 1 + x ^ { 16 } \right)$. Then $y ^ { \prime } - y ^ { \prime \prime }$ at $x = - 1$ is equal to
(1) 976
(2) 464
(3) 496
(4) 944

If $f(x) = x^2 + g'(1)x + g''(2)$ and $g(x) = f(1)x^2 + xf'(x) + f''(x)$, then the value of $f(4) - g(4)$ is equal to $\_\_\_\_$.

$\lim _ { x \rightarrow 0 } \frac { e ^ { 2 \sin x } - 2 \sin x - 1 } { x ^ { 2 } }$
(1) is equal to $-1$
(2) does not exist
(3) is equal to 1
(4) is equal to 2

Let $f : ( - \infty , \infty ) - \{ 0 \} \rightarrow \mathbb { R }$ be a differentiable function such that $f ^ { \prime } ( 1 ) = \lim _ { a \rightarrow \infty } a ^ { 2 } f \left( \frac { 1 } { a } \right)$. Then $\lim _ { a \rightarrow \infty } \frac { a ( a + 1 ) } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { a } \right) + a ^ { 2 } - 2 \log _ { e } a$ is equal to
(1) $\frac { 3 } { 2 } + \frac { \pi } { 4 }$
(2) $\frac { 3 } { 4 } + \frac { \pi } { 8 }$
(3) $\frac { 3 } { 8 } + \frac { \pi } { 4 }$
(4) $\frac { 5 } { 2 } + \frac { \pi } { 8 }$

If $\lim _ { x \rightarrow 0 } \frac { 3 + \alpha \sin x + \beta \cos x + \log _ { e } ( 1 - x ) } { 3 \tan ^ { 2 } x } = \frac { 1 } { 3 }$, then $2 \alpha - \beta$ is equal to :
(1) 2
(2) 7
(3) 5
(4) 1

If $a = \lim _ { x \rightarrow 0 } \frac { \sqrt { 1 + \sqrt { 1 + x ^ { 4 } } } - \sqrt { 2 } } { x ^ { 4 } }$ and $b = \lim _ { x \rightarrow 0 } \frac { \sin ^ { 2 } x } { \sqrt { 2 } - \sqrt { 1 + \cos x } }$, then the value of $a b ^ { 3 }$ is:
(1) 36
(2) 32
(3) 25
(4) 30

Suppose for a differentiable function $h , h ( 0 ) = 0 , h ( 1 ) = 1$ and $h ^ { \prime } ( 0 ) = h ^ { \prime } ( 1 ) = 2$. If $\mathrm { g } ( x ) = h \left( \mathrm { e } ^ { x } \right) \mathrm { e } ^ { h ( x ) }$, then $g ^ { \prime } ( 0 )$ is equal to:
(1) 5
(2) 4
(3) 8
(4) 3

Let $g(x)$ be a linear function and $f(x) = \begin{cases} g(x), & x \leq 0 \\ \frac { 1 + x } { 2 + x } , & x > 0 \end{cases}$, is continuous at $x = 0$. If $f'(1) = f(-1)$, then the value of $g(3)$ is
(1) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { e^{1/3} }$
(2) $\frac { 1 } { 3 } \log _ { e } \frac { 4 } { 9 } + 1$
(3) $\log _ { e } \frac { 4 } { 9 } - 1$
(4) $\log _ { e } \frac { 4 } { 9 e ^ { 1/3 } }$

Suppose $f ( x ) = \frac { \left( 2 ^ { x } + 2 ^ { - x } \right) \tan x \sqrt { \tan ^ { - 1 } \left( x ^ { 2 } - x + 1 \right) } } { \left( 7 x ^ { 2 } + 3 x + 1 \right) ^ { 3 } }$. Then the value of $f ^ { \prime } ( 0 )$ is equal to
(1) $\pi$
(2) 0
(3) $\sqrt { \pi }$
(4) $\frac { \pi } { 2 }$

Let $f ( x ) = x ^ { 3 } + x ^ { 2 } f ^ { \prime } ( 1 ) + x f ^ { \prime \prime } ( 2 ) + f ^ { \prime \prime \prime } ( 3 ) , x \in R$. Then $f ^ { \prime } ( 10 )$ is equal to

Let $\mathrm { a } > 0$ be a root of the equation $2 x ^ { 2 } + x - 2 = 0$. If $\lim _ { x \rightarrow \frac { 1 } { \mathrm { a } } } \frac { 16 \left( 1 - \cos \left( 2 + x - 2 x ^ { 2 } \right) \right) } { ( 1 - \mathrm { a } x ) ^ { 2 } } = \alpha + \beta \sqrt { 17 }$, where $\alpha , \beta \in Z$, then $\alpha + \beta$ is equal to $\_\_\_\_$

The value of $\lim _ { x \rightarrow 0 } 2 \left( \frac { 1 - \cos x \sqrt { \cos 2 x } \sqrt [ 3 ] { \cos 3 x } \ldots \ldots \sqrt [ 10 ] { \cos 10 x } } { x ^ { 2 } } \right)$ is $\_\_\_\_$